problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
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6. Given $a_{1}, a_{2}, \cdots$ is a geometric sequence with the first term $a_{1}=a\left(a \in \mathbf{Z}_{+}\right)$ and common ratio $r\left(r \in \mathbf{Z}_{+}\right)$. Suppose
$$
\log _{4} a_{2}+\log _{4} a_{3}+\cdots+\log _{4} a_{12}=2013 .
$$
Then the number of ordered pairs $(a, r)$ that satisfy the condition... | 6. 62.
From the given, we have $a_{n}=a r r^{n-1}$, substituting into the given equation yields
$$
a^{11} r^{66}=2^{4026} \Rightarrow a r^{6}=2^{366} \text{. }
$$
Let $a=2^{x}, r=2^{y}(x, y \in \mathrm{N})$. Then $x+6y=366$.
Thus, the number of pairs satisfying the condition is 62.
_ 171
2. $\frac{1}{3}$.
By the uni... | 62 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Question 3 Given $a b c=-1, a^{2} b+b^{2} c+c^{2} a=t$, $\frac{a^{2}}{c}+\frac{b}{c^{2}}=1$. Try to find the value of $a b^{5}+b c^{5}+c a^{5}$. | 【Analysis】Using $a b c=-1$ can make common substitutions
$$
a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x} \text {. }
$$
This problem involves three letters, and can also be solved by the method of elimination.
Solution 1 Let $a=-\frac{x}{y}, b=-\frac{y}{z}, c=-\frac{z}{x}$.
$$
\begin{array}{l}
\text { Then } \frac{a^... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $\left\{a_{n}\right\}$ be a geometric sequence, and each term is greater than 1. Then
$\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=$ $\qquad$ [2]
(2012, Zhejiang Province High School Mathematics Competition) | When the common ratio $q=1$,
$$
\begin{array}{l}
a_{n}=a_{1}, \\
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=2011 .
\end{array}
$$
When the common ratio $q \neq 1$,
$$
\begin{array}{l}
\lg a_{1} \cdot \lg a_{2012}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}} \\
=\frac{\lg a_{1}... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given positive integers $a_{1}, a_{2}, \cdots, a_{10}$ satisfy
$$
\frac{a_{j}}{a_{i}}>\frac{2}{3}(1 \leqslant i \leqslant j \leqslant 10) \text {. }
$$
Then the minimum possible value of $a_{10}$ is $\qquad$ . | 2. 92.
From $a_{1} \geqslant 1, a_{2}>\frac{3}{2} a_{1} \geqslant \frac{3}{2}\left(a_{2} \in \mathbf{N}_{+}\right)$, we get $a_{2} \geqslant 2$.
Similarly, $a_{3}>\frac{3}{2} a_{2} \geqslant 3, a_{3} \geqslant 4$;
$$
\begin{array}{l}
a_{4}>\frac{3}{2} a_{3} \geqslant 6, a_{4} \geqslant 7 ; \\
a_{5}>\frac{3}{2} a_{4} \... | 92 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $\tan \alpha+\tan \beta+\tan \gamma=\frac{17}{6}$, $\cot \alpha+\cot \beta+\cot \gamma=-\frac{4}{5}$, $\cot \alpha \cdot \cot \beta+\cot \beta \cdot \cot \gamma+\cot \gamma \cdot \cot \alpha=-\frac{17}{5}$. Then $\tan (\alpha+\beta+\gamma)=$ $\qquad$ | 3. 11 .
Let $\tan \alpha=x, \tan \beta=y, \tan \gamma=z$.
Then the given equations can be written as
$$
\begin{array}{l}
x+y+z=\frac{17}{6}, \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{4}{5}, \\
\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-\frac{17}{5} . \\
\text { From (1) } \div \text { (3) we get } \\
x y z=-\frac{... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, \\
a_{n+2}=\frac{2(n+1)}{n+2} a_{n+1}-\frac{n}{n+2} a_{n}(n=1,2, \cdots) .
\end{array}
$$
If $a_{m}>2+\frac{2011}{2012}$, then the smallest positive integer $m$ is . $\qquad$ | 8.4025.
$$
\begin{array}{l}
\text { Given } a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} \\
\begin{array}{l}
\Rightarrow a_{n}-a_{n-1}=\frac{n-2}{n}\left(a_{n-1}-a_{n-2}\right) \\
=\frac{n-2}{n} \cdot \frac{n-3}{n-1}\left(a_{n-2}-a_{n-3}\right)=\cdots \\
= \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \cdots \fr... | 4025 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. (16 points) Given an integer $n(n \geqslant 3)$, let $f(n)$ be the minimum number of elements in a subset $A$ of the set $\left\{1,2, \cdots, 2^{n}-1\right\}$ that satisfies the following two conditions:
(i) $1 \in A, 2^{n}-1 \in A$;
(ii) Each element in subset $A$ (except 1) is the sum of two (possibly the same) e... | 12. (1) Let set $A \subseteq\left\{1,2, \cdots, 2^{3}-1\right\}$, and $A$ satisfies (i) and (ii). Then $1 \in A, 7 \in A$.
Since $\{1, m, 7\}(m=2,3, \cdots, 6)$ does not satisfy (ii), hence $|A|>3$.
Also, $\{1,2,3,7\},\{1,2,4,7\},\{1,2,5,7\}$,
$\{1,2,6,7\},\{1,3,4,7\},\{1,3,5,7\}$,
$\{1,3,6,7\},\{1,4,5,7\},\{1,4,6,7\}... | 108 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $a, b, c$ satisfy
$$
a+b+c=a^{2}+b^{2}+c^{2} \text {. }
$$
Then the maximum value of $a+b+c$ is $\qquad$ | 4.3.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
3(a+b+c) \\
=\left(1^{2}+1^{2}+1^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)
\end{array}
$$
$$
\begin{array}{l}
\geqslant(a+b+c)^{2} \\
\Rightarrow a+b+c \leqslant 3 .
\end{array}
$$
Equality holds if and only if \(a=b=c=1\). | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Arrange the positive integers whose sum of digits is 5 in ascending order to form a sequence. Then 2012 is the $\qquad$th term of this sequence. | 7.38.
To represent 5 as the sum of no more than four positive integers, there are six methods, that is
$$
\begin{array}{l}
5=1+4=2+3=1+1+3 \\
=1+2+2=1+1+1+2 .
\end{array}
$$
When filling them into a $1 \times 4$ grid, positions that are not filled are supplemented with 0. Then $\{5\}$ has 3 ways of filling; $\{1,4\}$... | 38 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $18^{2}=324, 24^{2}=576$, they are formed by the permutation of two consecutive digits $2,3,4$ and $5,6,7$ respectively; and $66^{2}=4356$ is formed by the permutation of four consecutive digits $3, 4, 5, 6$. Then the next such square number is $\qquad$
| 8.5476.
For any square number, its last digit can only be $0, 1, 4, 5, 6, 9$, and
$$
\begin{array}{l}
(10 a)^{2}=100 a^{2}, \\
(10 a+5)^{2}=100 a^{2}+100 a+25, \\
(10 a \pm 4)^{2}=100 a^{2} \pm 80 a+16 ; \\
(10 a \pm 1)^{2}=100 a^{2} \pm 20 a+1, \\
(10 a \pm 3)^{2}=100 a^{2} \pm 60 a+9 ; \\
(10 a \pm 2)^{2}=100 a^{2} ... | 5476 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $4^{a}-3 a^{b}=16, \log _{2} a=\frac{a+1}{b}$. Then $a^{b}=$ $\qquad$ . | - 1. 16.
From $\log _{2} a=\frac{a+1}{b} \Rightarrow a^{b}=2^{a+1}$.
Substituting into $4^{a}-3 a^{b}=16$, we get
$2^{2 a}-6 \times 2^{a}-16=0$.
Solving, we get $2^{a}=8$ or -2 (discard).
Therefore, $a^{b}=2^{a+1}=16$. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that the base edge length of a regular tetrahedron is 6, and the side edge is 4. Then the radius of the circumscribed sphere of this regular tetrahedron is $\qquad$ | 3. 4 .
From the problem, we know that the distance from point $A$ to the base $B C D$ is 2, and the radius of its circumscribed sphere is $R$. Then
$$
R^{2}-12=(R-2)^{2} \Rightarrow R=4 \text {. }
$$ | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $S_{4} \leqslant 4, S_{5} \geqslant 15$, then the minimum value of $a_{4}$ is $\qquad$ . | 4. 7 .
Let the common difference be $d$. From the given conditions, we have
$$
\begin{array}{l}
2 a_{4} \leqslant 2+3 d, d \leqslant a_{4}-4 \\
\Rightarrow 2 a_{4} \leqslant 2+3 d \leqslant 2+3\left(a_{4}-3\right) \\
\Rightarrow a_{4} \geqslant 7 .
\end{array}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (16 points) Let the side lengths opposite to two interior angles of $\triangle A B C$ be $a, b, c$ respectively, and $a+b+c=16$. Find
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$
the value. | $$
\begin{array}{l}
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 4 R^{2}\left(\sin ^{2} B \cdot \cos ^{2} \frac{C}{2}+\sin ^{2} C \cdot \cos ^{2} \frac{B}{2}+\right. \\
\left.2 \sin B \cdot \sin C \cdot \cos \frac{B}{2} \cdot \cos \frac... | 64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 As shown in Figure 3, in the acute triangle $\triangle ABC$, it is known that $BE \perp AC$ at point $E$, $CD \perp AB$ at point $D$, $BC=25$, $CE=7$, $BD=15$. If $BE$ and $CD$ intersect at point $H$, connect $DE$, and construct a circle with $DE$ as the diameter, which intersects $AC$ at another point $F$. F... | Solve for DF. From the given conditions, we have
$$
\begin{array}{l}
\cos B=\frac{3}{5}, \cos C=\frac{7}{25} \\
\Rightarrow \sin B=\frac{4}{5}, \sin C=\frac{24}{25} \\
\Rightarrow \sin A=\sin B \cdot \cos C+\sin C \cdot \cos B=\frac{4}{5}=\sin B \\
\Rightarrow \angle A=\angle B \Rightarrow AC=BC=25 .
\end{array}
$$
Fr... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
16. Let $P$ be any point on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ other than the endpoints of the major axis, $F_{1}$ and $F_{2}$ be the left and right foci respectively, and $O$ be the center. Then
$\left|P F_{1}\right|\left|P F_{2}\right|+|O P|^{2}=$ $\qquad$ . | 16. 25.
According to the definition of an ellipse and the cosine rule, the solution can be found. | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let two ellipses be
$$
\frac{x^{2}}{t^{2}+2 t-2}+\frac{y^{2}}{t^{2}+t+2}=1
$$
and $\frac{x^{2}}{2 t^{2}-3 t-5}+\frac{y^{2}}{t^{2}+t-7}=1$
have common foci. Then $t=$ $\qquad$ . | 5.3.
Given that the two ellipses have a common focus, we have
$$
\begin{array}{l}
t^{2}+2 t-2-\left(t^{2}+t+2\right) \\
=2 t^{2}-3 t-5-\left(t^{2}+t-7\right) \\
\Rightarrow t=3 \text { or } 2 \text { (rejected). }
\end{array}
$$
Therefore, $t=3$. | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 (An Ancient Chinese Mathematical Problem) Emperor Taizong of Tang ordered the counting of soldiers: if 1,001 soldiers make up one battalion, then one person remains; if 1,002 soldiers make up one battalion, then four people remain. This time, the counting of soldiers has at least $\qquad$ people. | Let the first troop count be 1001 people per battalion, totaling $x$ battalions, then the total number of soldiers is $1001 x + 1$ people; let the second troop count be 1002 people per battalion, totaling $y$ battalions, then the total number of soldiers is $1002 y + 4$ people.
From the equality of the total number of ... | 1000000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Calculate:
$$
\sum_{k=0}^{2013}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2014}^{k}}=
$$
$\qquad$ | 5. 0 .
Let $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1) \frac{1}{\mathrm{C}_{2 n}^{k}}$.
And $\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n}-k-1}$, so $a_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{\mathrm{C}_{2 n}^{2 n... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Let $a, b, c > 0$, and $a+b+c=3$. Find the maximum value of $a^{2} b+b^{2} c+c^{2} a+a b c$.
| Let's assume $a \leqslant b \leqslant c$ or $c \leqslant b \leqslant a$.
Then $c(b-a)(b-c) \leqslant 0$
$$
\begin{aligned}
\Rightarrow & b^{2} c+c^{2} a \leqslant a b c+c^{2} b \\
\Rightarrow & a^{2} b+b^{2} c+c^{2} a+a b c \\
& \leqslant a^{2} b+c^{2} b+2 a b c \\
& =b(a+c)^{2}=b(3-b)^{2} \\
& \leqslant \frac{1}{2}\le... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50) Find the maximum value of $n$ such that there are $n$ points in the plane, where among any three points, there must be two points whose distance is 1. | If there exist $n(n \geqslant 8)$ points satisfying the conditions of the problem, let $V=\left\{v, v_{1}, v_{2}, \cdots, v_{7}\right\}$ represent any eight of these points. When and only when the distance between two points is 1, connect an edge between these two points, forming a graph $G$.
If there exists a point (... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given $x, y$ are positive real numbers, $n \in \mathrm{N}$, and $n \geqslant 2$. Prove:
$$
\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}}+\sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}} \geqslant 4 \text {. }
$$ | $$
\begin{array}{l}
\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}}+\sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}} \\
\geqslant 2 \sqrt{\sqrt[n]{\frac{x+\left(2^{n}-1\right) y}{x}} \cdot \sqrt[n]{\frac{y+\left(2^{n}-1\right) x}{y}}} \\
=2 \sqrt[2 n]{\frac{\left(2^{n}-1\right)\left(x^{2}+y^{2}\right)+\left(2^{2 n}-2^{n+1}+2\... | 4 | Inequalities | proof | Yes | Yes | cn_contest | false |
Example 3 Mother's Day is coming, and Xiao Hong, Xiao Li, and Xiao Meng went to the flower shop to buy flowers for their mothers. Xiao Hong bought 3 roses, 7 carnations, and 1 lily, and paid 14 yuan; Xiao Li bought 4 roses, 10 carnations, and 1 lily, and paid 16 yuan; Xiao Ying bought 2 stems of each of the three types... | Let the unit prices of roses, carnations, and lilies be $x$ yuan, $y$ yuan, and $\sqrt{z}$ yuan, respectively. Then,
$$
\left\{\begin{array}{l}
3 x+7 y+z=14, \\
4 x+10 y+z=16 .
\end{array}\right.
$$
Eliminating $z$ gives
$$
x=2-3 y \text{. }
$$
Substituting equation (2) into equation (1) gives
$$
z=8+2 y \text{. }
$$... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 In the tetrahedron $A-B C D$, it is known that
$$
\begin{array}{l}
\angle A C B=\angle C B D \\
\angle A C D=\angle A D C=\angle B C D=\angle B D C=\theta,
\end{array}
$$
and $\cos \theta=\frac{\sqrt{10}}{10}$.
If the length of edge $A B$ is $6 \sqrt{2}$, then the volume of this pyramid is
$\qquad$ | Solve As shown in Figure 2, from the problem, we know $\triangle A C D \cong \triangle B C D$,
and $\square$
$$
\begin{array}{l}
A C=A D \\
=B C=B D .
\end{array}
$$
Then $\triangle A C D$
$\cong \triangle B C D$
$\cong \triangle C A B$
$\cong \triangle D A B$.
Therefore, $\angle A C B$
$$
\begin{array}{l}
\because \a... | 144 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let the functions $f(x)=\ln x, g(x)=\frac{1}{2} x^{2}$. If $x_{1}>x_{2}>0$, for what value of $m(m \in \mathbf{Z}, m \leqslant 1)$ is it always true that
$$
m\left(g\left(x_{1}\right)-g\left(x_{2}\right)\right)>x_{1} f\left(x_{1}\right)-x_{2} f\left(x_{2}\right)
$$
holds. | Introduce an auxiliary function
$$
t(x)=m g(x)-x f(x)=\frac{m}{2} x^{2}-x \ln x \quad (x>0) \text {. }
$$
By the problem, $x_{1}>x_{2}>0$.
Therefore, if the original inequality always holds for $x>0$, i.e., the function $t(x)$ is monotonically increasing, then
$$
t^{\prime}(x)=m x-\ln x-1 \geqslant 0
$$
always holds.... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Two boxes of candy have a total of 176 pieces. 16 pieces are taken from the second box and placed into the first box, at which point, the number of pieces of candy in the first box is 31 more than $m($ an integer $m>1)$ times the number of pieces of candy in the second box. Then, the first box originally had ... | Let the first box originally contain $x$ candies, and the second box originally contain $y$ candies.
According to the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
x+y=176, \\
x+16=m(y-16)+31 .
\end{array}\right.
$$
Rearranging, we get
$$
x+16=m(176-16-x)+31 \text {, }
$$
which simplifies to ... | 131 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the system of inequalities about $x$
$$
\left\{\begin{array}{l}
3 x-3 \geqslant 6 x+a, \\
x \geqslant 1
\end{array}\right.
$$
the solution is $1 \leqslant x \leqslant 3$. Then $a=$ | $=, 1 .-12$.
From the given, we have $1 \leqslant x \leqslant \frac{1}{3}(-a-3)$. Then $\frac{1}{3}(-a-3)=3 \Rightarrow a=-12$. | -12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $x$ and $y$ be two distinct non-negative integers, and satisfy $x y + 2x + y = 13$. Then the minimum value of $x + y$ is $\qquad$ | 3. 5 .
From the problem, we know that $(x+1)(y+2)=15$.
$$
\begin{array}{l}
\text { Then }(x+1, y+2) \\
=(15,1),(5,3),(3,5),(1,15) \\
\Rightarrow(x, y)=(14,-1),(4,1),(2,3),(0,13) .
\end{array}
$$
Therefore, the minimum value of $x+y$ is 5. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given three points $A, B, C$ in a plane satisfying
$$
|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=5,|\overrightarrow{C A}|=6 \text {. }
$$
Then the value of $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}$ is ( ). | 5. C.
From the cosine theorem, we know
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}=-\overrightarrow{B A} \cdot \overrightarrow{B C} \\
=-\frac{|\overrightarrow{B A}|^{2}+|\overrightarrow{B C}|^{2}-|\overrightarrow{A C}|^{2}}{2} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\overrightarrow{B ... | -35 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. In the arithmetic sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{20} \simeq \frac{1}{a}, a_{201}=\frac{1}{b}, a_{2012}=\frac{1}{c} \text {. }
$$
Then $1992 a c-1811 b c-181 a b=$ | 12. 0 .
Let the common difference of the arithmetic sequence be $d$. Then, according to the problem, we have
$$
\begin{array}{l}
a_{201}-a_{20}=\frac{a-b}{a b}=181 d, \\
a_{2012}-a_{201}=\frac{b-c}{b c}=1811 d, \\
a_{2012}-a_{20}=\frac{a-c}{a c}=1992 d .
\end{array}
$$
Therefore, $1992 a c-1811 b c-181 a b$
$$
=\frac... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Given constants $a, b$ satisfy $a, b>0, a \neq 1$, and points $P(a, b), Q(b, a)$ are both on the curve $y=\cos (x+c)$, where $c$ is a constant. Then $\log _{a} b=$ $\qquad$ | 13. 1 .
Given points $P(a, b), Q(b, a)$ are both on the curve
$$
y=\cos (x+c)
$$
we know
$$
\begin{array}{l}
a-b=\cos (b+c)-\cos (a+c) \\
=2 \sin \left(\frac{a+b+2 c}{2}\right) \cdot \sin \frac{a-b}{2} .
\end{array}
$$
Without loss of generality, assume $a \geqslant b$.
If $a>b$, then
$$
\left|\sin \frac{a+b+2 c}{2}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Every day at 5 PM when school is over, Xiao Ming's father always drives from home to pick him up on time and take him back. One day, the school dismissed an hour early, and Xiao Ming walked home by himself. On the way, he met his father who was coming to pick him up, and as a result, they arrived home 20 minutes ear... | 二、1.50 minutes.
As shown in Figure 6, Xiao Ming
starts walking home from point $A$
and meets the car coming to pick him up
at point $C$. As a result, the car returns from $C$ to $B$ 20 minutes earlier than usual. This indicates that the car takes 20 minutes to travel from $C$ to $A$ and back to $C$.
Therefore, the car ... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
2. Let real numbers $x, y, z, w$ satisfy $x \geqslant y \geqslant z \geqslant w \geqslant 0$, and $5 x+4 y+3 z+6 w=100$. Denote the maximum value of $x+y+z+w$ as $a$, and the minimum value as $b$. Then $a+b=$ $\qquad$ | 2. 45 .
From $x \geqslant y \geqslant z \geqslant w \geqslant 0$, we know
$$
\begin{array}{l}
100=5 x+4 y+3 z+6 w \geqslant 4(x+y+z+w) \\
\Rightarrow x+y+z+w \leqslant 25 .
\end{array}
$$
When $x=y=z=\frac{25}{3}, w=0$, the equality holds.
$$
\begin{array}{l}
\text { Also } 100=5 x+4 y+3 z+6 w \leqslant 5(x+y+z+w) \\... | 45 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $m$ is an integer, the equation
$$
x^{2}-m x+3-n=0
$$
has two distinct real roots, the equation
$$
x^{2}+(6-m) x+7-n=0
$$
has two equal real roots, and the equation
$$
x^{2}+(4-m) x+5-n=0
$$
has no real roots.
$$
\text { Then }(m-n)^{2013}=
$$ | 3. According to the problem, we have
$$
\left\{\begin{array}{l}
m^{2}-4(3-n)>0, \\
(6-m)^{2}-4(7-n)=0, \\
(4-m)^{2}-4(5-n)\frac{5}{3}, m<3 \text {. }
\end{array}\right.
$$
Then $m=2$. Consequently, $n=3$.
Therefore, $(m-n)^{2013}=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) In a dormitory of a school, there are several students. On New Year's Day, each student in the dormitory gives a greeting card to every other student, and each student also gives a greeting card to each dormitory administrator, who in turn gives a card back to each student. In this way, a total of 51 g... | Let there be $x$ students and $y$ administrators living in the dormitory $\left(x, y \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
x(x-1)+x y+y=51 \\
\Rightarrow y=\frac{51+x-x^{2}}{x+1}=\frac{49}{x+1}-x+2 .
\end{array}
$$
Since $x, y$ are positive integers, we know that $(x, y)=(6,3)$.
Therefore, there are 6 s... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. In $\triangle A B C$, it is known that $O$ is the circumcenter, the three altitudes $A D, B E, C F$ intersect at point $H$, line $E D$ intersects $A B$ at point $M$, and $F D$ intersects $A C$ at point $N$. Then $\overrightarrow{O H} \cdot \overrightarrow{M N}=$ $\qquad$ | 3. 0 .
It is known that, $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{O H}$.
Then $\overrightarrow{O H} \cdot \overrightarrow{M N}=(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \cdot \overrightarrow{M N}$
$$
\begin{aligned}
= & \overrightarrow{O A} \cdot(\overright... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the real number pair $(x, y)$ satisfies the equation $(x-2)^{2}+y^{2}=3$, let the minimum and maximum values of $\frac{y}{x}$ be $m$ and $n$ respectively. Then $m+n=$ | $\begin{array}{l}\text { II. 1.0. } \\ \text { Let } y=t x \text {. Then }\left(1+t^{2}\right) x^{2}-4 x+1=0 \text {. } \\ \text { By } \Delta=(-4)^{2}-4\left(1+t^{2}\right) \geqslant 0 \\ \Rightarrow-\sqrt{3} \leqslant t \leqslant \sqrt{3} \\ \Rightarrow m=-\sqrt{3}, n=\sqrt{3} \\ \Rightarrow m+n=0 \text {. }\end{arra... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. After rotating any positive integer by $180^{\circ}$, some interesting phenomena can be observed, such as 808 still being 808 after a $180^{\circ}$ rotation, 169 becoming 691 after a $180^{\circ}$ rotation, and 37 not being a number after a $180^{\circ}$ rotation. Then, among all five-digit numbers, the number of fi... | 2. 60.
Among the ten digits from $0$ to $9$, $(0,0)$, $(1,1)$, $(8,8)$, and $(6,9)$ can be placed in the symmetric positions at the beginning and end of a five-digit number. When rotated $180^{\circ}$, the resulting number is the same as the original number, while other digits cannot appear in the five-digit number. S... | 60 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 2, in
rhombus $A B C D$, it is
known that $\angle A B C=60^{\circ}$, line
$E F$ passes through point $D$, and
intersects the extensions of
$B A$ and $B C$ at points $E$ and $F$, respectively. $M$
is the intersection of $C E$ and $A F$. If $C M=4, E M=5$, then $C A=$ $\qquad$ | 3. 6 .
It is easy to prove $\triangle E A D \backsim \triangle D C F$.
$$
\begin{array}{l}
\text { Therefore, } \frac{E A}{A D}=\frac{D C}{C F} \Rightarrow \frac{E A}{A C}=\frac{A C}{C F} \\
\Rightarrow \triangle E A C \backsim \triangle A C F \\
\Rightarrow \angle A E C=\angle C A F \\
\Rightarrow C A^{2}=C E \cdot C... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Let $x, y (x>y)$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, satisfying $x-y \geqslant \frac{xy}{31}$. Find the maximum value of the number of elements $n$ in this set of natural numbers. | Let's assume $a_{1}1 \\
\Rightarrow d_{6} \geqslant 2 \Rightarrow a_{7}=a_{6}+d_{6} \geqslant 8 \\
\Rightarrow d_{7} \geqslant \frac{8^{2}}{31-8}>2 \Rightarrow d_{7} \geqslant 3 . \\
\Rightarrow a_{8}=a_{7}+d_{7} \geqslant 11 \\
\Rightarrow d_{8} \geqslant \frac{11^{2}}{31-11}>6 \Rightarrow d_{8} \geqslant 7 \\
\Righta... | 10 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that the sum of the first $n(n>1)$ terms of an arithmetic sequence is 2013, the common difference is 2, and the first term is an integer. Then the sum of all possible values of $n$ is . $\qquad$ | 3. 2975 .
Let the first term of the sequence be $a_{1}$, and the common difference $d=2$. Then
$$
\begin{array}{l}
S_{n}=n a_{1}+n(n-1) \\
=n\left(a_{1}+n-1\right)=2013 .
\end{array}
$$
Also, $2013=3 \times 11 \times 61$, and $n$ is a divisor of 2013, so the sum of all possible values of $n$ is
$$
(1+3) \times(1+11) ... | 2975 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $a, b, x \in \mathbf{N}_{+}$, and $a \leqslant b$. $A$ is the solution set of the inequality
$$
\lg b - \lg a < \lg x < \lg b + \lg a
$$
It is known that $|A|=50$. When $ab$ takes its maximum possible value,
$$
\sqrt{a+b}=
$$ | 4. 6 .
It is easy to know, $\frac{b}{a}<x<a b, a \neq 1$.
Therefore, $a \geqslant 2$,
$$
\begin{array}{l}
50 \geqslant a b-\frac{b}{a}-1=a b\left(1-\frac{1}{a^{2}}\right)-1 \geqslant \frac{3}{4} a b-1 \\
\Rightarrow a b \leqslant 68 .
\end{array}
$$
Upon inspection, when and only when $a=2, b=34$, the equality holds.... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The function defined on the domain $R$
$$
f(x)=|\lg | x-2||-1 \text{. }
$$
If $b<0$, then the equation concerning $x$
$$
f^{2}(x)+b f(x)=0
$$
has $\qquad$ distinct real roots. | 5. 8 .
From the problem, we know that the graph of $y=\lg x$ is first symmetrical about the $y$-axis, then the part below the $x$-axis is flipped up, followed by a rightward shift of 2 units and a downward shift of 1 unit to form the graph of $f(x)$. The zeros of $g(x)=f^{2}(x)+b f(x)$ are the roots of the equation. I... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=p, a_{2}=p+1, \\
a_{n+2}-2 a_{n+1}+a_{n}=n-20,
\end{array}
$$
where $p$ is a given real number, and $n$ is a positive integer. Try to find the value of $n$ that minimizes $a_{n}$. | 10. Let $b_{n}=a_{n+1}-a_{n}$.
From the problem, we have $b_{n+1}-b_{n}=n-20$, and $b_{1}=1$. Then, $b_{n}-b_{1}=\sum_{i=1}^{n-1}\left(b_{i+1}-b_{i}\right)=\sum_{i=1}^{n-1}(i-20)$.
Thus, $b_{n}=\frac{(n-1)(n-40)}{2}+1$.
Also, $a_{3}=a_{2}+b_{2}=p-17<a_{1}<a_{2}$, so when the value of $a_{n}$ is the smallest, we have $... | 40 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given $x, y \in \mathbf{N}_{+}$. Find the minimum value of $\sqrt{512^{x}-7^{2}-1}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $z=512^{x}-7^{2 y-1}$.
Obviously, $z \geqslant 0$, and
$z \equiv 1(\bmod 7), z \equiv 1(\bmod 8)$.
(1) $x=2 x_{1}-1$ is an odd number.
Then $z \equiv(-1)^{x}-1 \equiv-2 \equiv 1(\bmod 3)$.
Thus, $z \equiv 1(\bmod 3 \times 7 \times 8)$.
Let $z=1$, i.e., $512^{2 x_{1}-1}-7^{2 y-1}=1$. Then $7^{2 y-1}=512^{2 x_{1}-1}... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $z$ be a complex number with modulus 2. Then the sum of the maximum and minimum values of $\left|z-\frac{1}{z}\right|$ is $\qquad$ [2] | Given $|z|=2$, we know
$$
\begin{array}{l}
|z+1|^{2}=(z+1)(\bar{z}+1) \\
=z \bar{z}+z+\bar{z}+1=5+2 \operatorname{Re} z, \\
|z-1|^{2}=(z-1)(\bar{z}-1) \\
=z \bar{z}-z-\bar{z}+1=5-2 \operatorname{Re} z .
\end{array}
$$
Therefore, $\left|z-\frac{1}{z}\right|=\left|\frac{z^{2}-1}{z}\right|$
$$
=\frac{|z+1||z-1|}{|z|}=\fr... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given
$$
A=\left\{z \mid z^{18}=1\right\} \text { and } B=\left\{\omega \mid \omega^{48}=1\right\}
$$
are sets of complex roots of unity,
$$
C=\{z w \mid z \in A, w \in B\}
$$
is also a set of complex roots of unity. How many distinct elements are there in the set $C$? ${ }^{[3]}$ | Notice that, $z=\cos \frac{2 k \pi}{18}+\mathrm{i} \sin \frac{2 k \pi}{18}(k \in \mathbf{Z})$ (18 distinct elements),
$\omega=\cos \frac{2 t \pi}{48}+\mathrm{i} \sin \frac{2 t \pi}{48}(t \in \mathbf{Z})$ (48 distinct elements),
$$
\begin{array}{l}
z \omega=\cos \frac{2 \pi(8 k+3 t)}{144}+\mathrm{i} \sin \frac{2 \pi(8 k... | 144 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For a natural number $n$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and their sum is 17. If there exists a unique $n$ such that $S_{n}$ is also an integer, find $n .{ }^{(4]}$ | Solve: Regarding $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$ as the modulus of the complex number $(2 k-1)+a_{k} \mathrm{i}$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}} \\
=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} \mathrm{i}\right| \\
\geqslant\left|\sum_{k=1}^{n}\left[(2 k-1)+a_{k} \mat... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In rectangle $A B C D$, it is known that $A B=5, B C=9$, points $E, F, G, H$ are on sides $A B, B C, C D, D A$ respectively, such that $A E=C G=3, B F=D H=4, P$ is a point inside the rectangle. If the area of quadrilateral $A E P H$ is 15, then the area of quadrilateral $P F C G$ is $\qquad$ | 4. 11.
As shown in Figure 3, let the distances from $P$ to $AB$ and $AD$ be $a$ and $b$, respectively. Then the distances from $P$ to $BC$ and $CD$ are $5-b$ and $9-a$, respectively. Given that $S_{\text {quadrilateral } AEPH}=\frac{1}{2}(3a+5b)=15$, we have
$$
\begin{array}{l}
S_{\text {quadrilateral PFCG }}=\frac{1}... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. The number of integers $n$ that make $n^{4}-3 n^{2}+9$ a prime number is $\qquad$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 5.4.
If $n=0$, then $n^{4}-3 n^{2}+9=9$ is not a prime number.
If $n>0$, then
$$
\begin{array}{l}
n^{4}-3 n^{2}+9=\left(n^{2}+3\right)^{2}-(3 n)^{2} \\
=\left(n^{2}-3 n+3\right)\left(n^{2}+3 n+3\right) .
\end{array}
$$
Notice that, $n^{2}+3 n+3>3$.
Therefore, $n^{2}-3 n+3=1$.
Solving this gives $n=1$ or 2 (correspond... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that the upper base, height, and lower base of a trapezoid are three consecutive positive integers, and these three numbers make the value of the polynomial $x^{3}-30 x^{2}+a x$ (where $a$ is a constant) also three consecutive positive integers in the same order. Then the area of this trapezoid is $\qquad$ | 7. 100 .
Let the upper base, height, and lower base be $n-1$, $n$, and $n+1$ respectively, and the other three consecutive integers be $m-1$, $m$, and $m+1$. Then,
$$
\begin{array}{l}
(n-1)^{3}-30(n-1)^{2}+a(n-1)=m-1, \\
n^{3}-30 n^{2}+a n=m, \\
(n+1)^{3}-30(n+1)^{2}+a(n+1)=m+1 . \\
\text { (1) }+(3)-2 \times(2) \text... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Arrange all positive integers that leave a remainder of 2 and 3 when divided by 4 in ascending order. Let $S_{n}$ denote the sum of the first $n$ terms of this sequence. Then $\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2012}}\right]$ $=$ $\qquad$ ([ $x]$ denotes the greatest integer no... | 8. 2025078.
Given that this sequence is
$$
2,3,6,7, \cdots, 4 n-2,4 n-1, \cdots \text {. }
$$
From $4 n-2+4 n-1=8 n-3$, we know
$$
\begin{array}{l}
S_{2 n}=5+13+\cdots+(8 n-3) \\
=4 n(n+1)-3 n=4 n^{2}+n . \\
\text { Also, } 4 n^{2}<S_{2 n}<(2 n+1)^{2} \text {, then } \\
2 n<\sqrt{S_{2 n}}<2 n+1 .
\end{array}
$$
Ther... | 2025078 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 A scientist stored the design blueprint of his time machine in a computer, setting the password to open the file as a permutation of $\{1,2, \cdots, 64\}$. He also designed a program that, when eight positive integers between $1 \sim 64$ are input each time, the computer will indicate the order (from left to ... | Let the password be denoted as $a_{1} a_{2} \cdots a_{n^{2}} (n=8)$.
First, write the numbers $1, 2, \cdots, n^{2}$ arbitrarily into an $n \times n$ grid (one number per cell). After the first $n$ operations, input each row of numbers once, and rearrange the numbers in each row from left to right according to the compu... | 45 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given any positive integer $a$, define the integer sequence $x_{1}, x_{2}$, $\cdots$, satisfying
$$
x_{1}=a, x_{n}=2 x_{n-1}+1(n \geqslant 1) .
$$
If $y_{n}=2^{x_{n}}-1$, determine the maximum integer $k$ such that there exists a positive integer $a$ for which $y_{1}, y_{2}, \cdots, y_{k}$ are all prime numbers. | 1. If $y_{i}$ is a prime number, then $x_{i}$ is also a prime number.
Otherwise, if $x_{i}=1$, then $y_{i}=1$ is not a prime number; if $x_{i}=m n($ integers $m, n>1)$, then
$\left(2^{m}-1\right) \mid\left(2^{x_{i}}-1\right)$, i.e., $x_{i}$ and $y_{i}$ are composite numbers.
Below, we prove by contradiction: For any ... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the set
$$
A=\{x \mid 5 x-a \leqslant 0, a \in \mathbf{N}\} \text {. }
$$
If $5 \in A \cap \mathbf{Z}$, then the minimum value of $a$ is | $$
-, 1.25
$$
From $A \left\lvert\,=\left(-\infty, \frac{a}{5}\right]\right.$, and $5 \in A \cap \mathbf{Z}$, we know
$$
\frac{a}{5} \geqslant 5 \Rightarrow a \geqslant 25 \text {. }
$$
Therefore, the minimum value of $a$ is 25. | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a function $f(x)$ defined on $\mathbf{R}$ that satisfies
$$
\begin{array}{l}
f(x+1)=f(-x), \\
f(x)=\left\{\begin{array}{ll}
1, & -1<x \leqslant 0 \\
-1, & 0<x \leqslant 1 .
\end{array}\right.
\end{array}
$$
Then $f(f(3.5))=$ $\qquad$ | 2. -1 .
From $f(x+1)=-f(x)$, we know $f(x+2)=f(x)$.
Then $f(3.5)=f(-0.5)=1$.
Therefore, $f(f(3.5))=f(1)=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the function
$$
y=a^{x+3}-2(a>0, a \neq 1)
$$
the graph always passes through a fixed point $A$. If point $A$ lies on the line
$$
\frac{x}{m}+\frac{y}{n}+1=0(m, n>0)
$$
then the minimum value of $3 m+n$ is | 5. 16 .
Note that the function
$$
y=a^{x+3}-2(a>0, a \neq 1)
$$
always passes through the fixed point $(-3,-1)$.
So point $A(-3,-1)$.
Then $-\frac{3}{m}-\frac{1}{n}+1=0 \Rightarrow 1=\frac{3}{m}+\frac{1}{n}$.
Thus, $3 m+n=(3 m+n)\left(\frac{3}{m}+\frac{1}{n}\right)$
$$
=10+\frac{3 n}{m}+\frac{3 m}{n} \geqslant 16 \te... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $P$ is a moving point on the line $l$:
$$
k x+y+4=0(k>0)
$$
$P A$ and $P B$ are the two tangents from $P$ to the circle $C$:
$$
x^{2}+y^{2}-2 y=0
$$
with points of tangency $A$ and $B$ respectively. If the minimum area of quadrilateral $P A C B$ is 2, then $k=$ $\qquad$ | 6.2.
$$
\begin{array}{l}
\text { Given } S_{\text {quadrilateral } P A C B}=P A \cdot A C=P A \\
=\sqrt{C P^{2}-C A^{2}}=\sqrt{C P^{2}-1},
\end{array}
$$
we know that the area is minimized when $|C P|$ is minimized, i.e., when $C P \perp l$. Also, $\sqrt{C P^{2}-1}=2$, then $C P=\sqrt{5}$. Using the point-to-line dist... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. (14 points) As shown in Figure 1, in the triangular prism $A B C-A_{1} B_{1} C_{1}$, it is known that the side edge $A A_{1} \perp$ plane $A B C$, and $\triangle A B C$ is an equilateral triangle with a side length of 2. $M$ is a point on $A A_{1}$, $A A_{1}=4, A_{1} M=1, P$ is a point on the edge $B C$, and the sh... | 10. (1) From $A A_{1} \perp$ plane $A B C$ and $\triangle A B C$ being an equilateral triangle, we know that all the side faces are congruent rectangles.
As shown in Figure 3, rotate the side face $B C_{1}$ by $120^{\circ}$ so that it lies in the same plane as the side face $A C_{1}$. Point $P$ moves to the position o... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. (15 points) As shown in Figure 2, in the Cartesian coordinate system, the equation of circle $\odot M$ is
$$
x^{2}+y^{2}+D x+E y+F=0,
$$
and the quadrilateral $A B C D$ inscribed in $\odot M$ has diagonals $A C$ and $B D$ that are perpendicular to each other, with $A C$ and $B D$ lying on the $x$-axis and $y$-axis... | 13. (1) Let $A(a, 0), C(c, 0)$.
From the problem, points $A$ and $C$ are on the negative and positive halves of the $x$-axis, respectively. Thus, $a c<0$.
When $y=0$, the equation becomes
$$
x^{2}+D x+F=0 \text{, }
$$
where the two roots of the equation are the $x$-coordinates of points $A$ and $C$.
Therefore, $x_{1}... | 64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
I. Fill in the Blanks (8 points each, total 64 points)
1. Among the positive integers less than 20, choose three different numbers such that their sum is divisible by 3. The number of different ways to choose these numbers is $\qquad$. | $-, 1.327$
$$
C_{6}^{3}+C_{6}^{3}+C_{7}^{3}+6 \times 6 \times 7=327
$$ | 327 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. In $\triangle A B C$, it is known that $\angle C=90^{\circ}, \angle B=$ $30^{\circ}, A C=2, M$ is the midpoint of side $A B$, and $\triangle A C M$ is folded along $C M$ so that the distance between points $A$ and $B$ is $2 \sqrt{2}$. Then the distance from point $M$ to plane $A B C$ is $\qquad$ | 3. 1 .
From the problem, we know that $M A=M B=M C=2$.
Therefore, the projection $O$ of $M$ on the plane $A B C$ is the circumcenter of $\triangle A B C$.
Given $A B=2 \sqrt{2}, A C=2, B C=2 \sqrt{3}$, we know that $A B^{2}+A C^{2}=B C^{2}$.
Thus, $O$ is the midpoint of the hypotenuse $B C$ of the right triangle $\tri... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given a geometric sequence $\left\{a_{n}\right\}$ with all terms being positive. If $2 a_{4}+a_{3}-2 a_{2}-a_{1}=8$, then the minimum value of $2 a_{8}+a_{7}$ is $-2 x-2$ $\qquad$ | 6. 54.
Let $\left\{a_{n}\right\}$ be a geometric sequence with common ratio $q(q>0)$. Then, according to the problem,
$$
\begin{array}{l}
2 a_{2} q^{2}+a_{1} q^{2}-\left(2 a_{2}+a_{1}\right)=8 \\
\Rightarrow\left(2 a_{2}+a_{1}\right)\left(q^{2}-1\right)=8 \\
\Rightarrow 2 a_{2}+a_{1}=\frac{8}{q^{2}-1},
\end{array}
$$
... | 54 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
\begin{array}{l}
{\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\cdots+\left[\log _{2} 2012\right]} \\
=
\end{array}
$$ | 8. 18084.
When $2^{k} \leqslant x<2^{k+1}$, then $\left[\log _{2} x\right]=k$.
Given $1024=2^{10}<2012<2^{11}=2048$, we know
$$
\begin{array}{l}
{\left[\log _{2} 1024\right]+\left[\log _{2} 1025\right]+\cdots+\left[\log _{2} 2012\right]} \\
=10 \times(2012-1023)=9890 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence ... | 18084 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $[x]$ denote the integer part of the real number $x$. Then $[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{49}]=(\quad)$.
(A) 146
(B) 161
(C) 210
(D) 365 | 4. C.
$$
\begin{array}{l}
{[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{49}]} \\
=1 \times 3+2 \times 5+3 \times 7+4 \times 9+5 \times 11+6 \times 13+7 \\
=210
\end{array}
$$ | 210 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Example 1 From the 205 positive integers $1,2, \cdots, 205$, what is the maximum number of integers that can be selected such that for any three selected numbers $a, b, c (a<b<c)$, we have
$$
a b \neq c ?^{[1]}
$$
(2005, (Casio Cup) National Junior High School Mathematics Competition) | Estimate first.
Since $14 \times 15=210>205$, then $14,15, \cdots$, 205 satisfy that for any three numbers $a 、 b 、 c(a<b<c)$, we have $a b \neq c$.
Because 1 multiplied by any number equals the number itself, $1,14,15, \cdots, 205$ satisfy the condition.
Therefore, there are $205-14+1+1=193$ numbers in total.
If we se... | 193 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2: 9 judges score 12 athletes participating in a bodybuilding competition. Each judge gives 1 point to the athlete they consider to be in 1st place, 2 points to the athlete in 2nd place, ..., and 12 points to the athlete in 12th place. The final scoring shows: the difference between the highest and lowest score... | Explanation: It is impossible for 9 judges to give 1 point to five or more athletes, because among five or more athletes, at least one athlete must be rated no less than 5 by a judge. However, according to the problem, each of these five athletes is rated no more than 4 by each judge, which is a contradiction.
Therefo... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 In $\triangle A B C$, it is known that $\angle A: \angle B: \angle C = 4: 2: 1, \angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively.
(1) Prove: $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$;
(2) Find the value of $\frac{(a+b-c)^{2}}{a^{2}+b^{2}+c^{2}}$. | (1) Proof As shown in Figure 3, construct the angle bisector of $\angle ABC$, intersecting the circumcircle of $\triangle ABC$ at point $M$, and construct the angle bisector of $\angle BAC$, intersecting the circumcircle of $\triangle ABC$ at point $N$.
Then $AM=MC=AB=c, AM \parallel BC$;
$$
CN=NB=AC=b, AB \parallel CN... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the equation in $x$
$$
(a-1) x^{2}+2 x-a-1=0
$$
has roots that are all integers. Then the number of integer values of $a$ that satisfy this condition is
$\qquad$. | 4.5.
When $a=1$, $x=1$.
When $a \neq 1$, it is easy to see that $x=1$ is an integer root of the equation.
Furthermore, from $1+x=\frac{2}{1-a}$ and $x$ being an integer, we know
$$
1-a= \pm 1, \pm 2 \text {. }
$$
Therefore, $a=-1,0,2,3$.
In summary, there are 5 integer values of $a$ that satisfy the condition. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let the quadratic function
$$
f(x)=a x^{2}+b x+c
$$
satisfy the following conditions:
(i) When $x$ is a real number, its minimum value is 0, and
$$
f(x-1)=f(-x-1)
$$
holds;
(ii) There exists a real number $m (m>1)$, such that there exists a real number $t$, as long as $1 \leqslant x \leqslant m$, t... | (1) In condition (ii), let $x=1$, we get $f(1)=1$.
From condition (i), we know that the quadratic function opens upwards and is symmetric about $x=-1$, so we can assume the quadratic function to be
$$
f(x)=a(x+1)^{2}(a>0) .
$$
Substituting $f(1)=1$ into the above equation, we get $a=\frac{1}{4}$.
Therefore, $f(x)=\fra... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. From the arithmetic sequence $2,5,8, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$ ـ. | 7.8.
First, let's take $x_{1}, x_{2}, \cdots, x_{k}$ from the known sequence such that
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1 .
$$
Let $y_{i}=\frac{x_{1} x_{2} \cdots x_{k}}{x_{i}}$. Then
$$
y_{1}+y_{2}+\cdots+y_{k}=x_{1} x_{2} \cdots x_{k} \text {. }
$$
It is easy to see that for any $n$, $x_{n... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (40 points) Given the sequence of real numbers $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{3}, a_{n+1}=2 a_{n}-\left[a_{n}\right],
$$
where $[x]$ denotes the greatest integer less than or equal to the real number $x$.
$$
\text { Find } \sum_{i=1}^{2012} a_{i} \text {. }
$$ | $$
\text { II. } a_{1}=\frac{1}{3}, a_{2}=\frac{2}{3}, a_{3}=\frac{4}{3}, a_{4}=\frac{5}{3} .
$$
By mathematical induction, it is easy to prove
$$
\left\{\begin{array}{l}
a_{2 k+1}=\frac{3 k+1}{3}, \\
a_{2 k+2}=\frac{3 k+2}{3} .
\end{array}\right.
$$
Then $a_{2 k+1}+a_{2 k+2}=2 k+1$.
Therefore, $\sum_{i=1}^{2012} a_{... | 1012036 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Can 2010 be written as the sum of squares of $k$ distinct positive integers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason. ${ }^{[2]}$
(2010, Beijing Middle School Mathematics Competition (Grade 8)) | Estimate the approximate range of $k$ first, then discuss by classification.
Let $p_{i}$ be a prime number.
If 2010 can be written as the sum of squares of $k$ prime numbers, then by
$$
\begin{array}{l}
2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}+17^{2}+19^{2}+23^{2}+29^{2} \\
=2397>2010,
\end{array}
$$
we know $k \leqslant... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Choose $k$ numbers from $1,2, \cdots, 2004$, such that among the chosen $k$ numbers, there are definitely three numbers that can form the side lengths of a triangle (the three numbers must be distinct). What is the minimum value of $k$ that satisfies this condition? | When selecting three numbers from 1 to 2004 to form the sides of a triangle, there are too many possibilities. Instead, let's approach it from the opposite direction and list all sets of three numbers that cannot form the sides of a triangle.
First, 1, 2, 3 cannot form the sides of a triangle. Adding 5, the set \(1, 2... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. A. Let $a=\sqrt[3]{3}, b$ be the fractional part of $a^{2}$. Then $(b+2)^{3}=$ $\qquad$ | ニ、6. A.9.
From $2<a^{2}<3$, we know $b=a^{2}-2=\sqrt[3]{9}-2$. Therefore, $(b+2)^{3}=(\sqrt[3]{9})^{3}=9$. | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. A. Given positive integers $a$, $b$, $c$ satisfy
$$
\begin{array}{l}
a+b^{2}-2 c-2=0, \\
3 a^{2}-8 b+c=0 .
\end{array}
$$
Then the maximum value of $a b c$ is $\qquad$ | 8. A. 2013.
The two equations are simplified and rearranged to get
$$
(b-8)^{2}+6 a^{2}+a=66 \text {. }
$$
Given that $a$ is a positive integer and $6 a^{2}+a \leqslant 66$, we have $1 \leqslant a \leqslant 3$. If $a=1$, then $(b-8)^{2}=59$, which has no positive integer solutions; if $a=2$, then $(b-8)^{2}=40$, whic... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
B. If $k$ numbers are chosen from 2, $, 8, \cdots, 101$ these 34 numbers, where the sum of at least two of them is 43, then the minimum value of $k$ is: $\qquad$ | B. 28.
Divide the 14 numbers less than 43 into the following seven groups: $(2,41),(5,38),(8,35),(11,32)$, $(14,29),(17,26),(20,23)$.
The sum of the two numbers in each group is 43. After selecting one number from each group and then taking all numbers greater than 43, a total of 27 numbers are selected. The sum of a... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. A. Xiaoming volunteered to sell pens at a stationery store one day. Pencils were sold at 4 yuan each, and ballpoint pens at 7 yuan each. At the beginning, it was known that he had a total of 350 pencils and ballpoint pens. Although he did not sell them all that day, his sales revenue was 2013 yuan. Then he sold at ... | 10. A. 207.
Let $x$ and $y$ represent the number of pencils and ballpoint pens sold, respectively. Then
\[
\begin{array}{l}
\left\{\begin{array}{l}
4 x+7 y=2013 ; \\
x+y=204
\end{array}\right.
\end{array}
\]
Thus, $y_{\text {min }}=207$, at which point, $x=141$. | 207 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
B. In the Cartesian coordinate system $x O y$, it is known that $O$ is the origin, point $A(10,100), B\left(x_{0}, y_{0}\right)$, where $x_{0} 、 y_{0}$ are integers, and points $O 、 A 、 B$ are not collinear. For all points $B$ that satisfy the above conditions, find the minimum area of $\triangle O A B$. | B. Draw a line parallel to the $x$-axis through point $B$, intersecting line $O A$ at point $C\left(\frac{y_{0}}{10}, y_{0}\right)$.
Thus, $B C=\left|x_{0}-\frac{y_{0}}{10}\right|$.
Since points $O$, $A$, and $B$ are not collinear, we have
$$
\begin{array}{l}
S_{\triangle O A B}=\frac{1}{2} B C \times 100=50\left|x_{0}... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. A. As shown in Figure 5, given that $A B$ is the diameter of $\odot O$, $C$ is a point on the circumference, and $D$ is a point on the line segment $O B$ (not at the endpoints), satisfying $C D \perp A B$ and $D E \perp C O$ at point $E$. If $C E = 10$, and the lengths of $A D$ and $D B$ are both positive integers,... | 12. A. Connect $A C$ and $B C$, then $\angle A C B=90^{\circ}$.
From $\mathrm{Rt} \triangle C D E \backsim \mathrm{Rt} \triangle C O D$, we know $C E \cdot C O=C D^{2}$.
From $\mathrm{Rt} \triangle A C D \backsim \mathrm{Rt} \triangle C B D$, we know $C D^{2}=A D \cdot B D$.
Therefore, $C E \cdot C O=A D \cdot B D$.
L... | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
14. A. If placing the positive integer $M$ to the left of the positive integer $m$ results in a new number that is divisible by 7, then $M$ is called the "magic number" of $m$ (for example, placing 86 to the left of 415 results in the number 86415, which is divisible by 7, so 86 is called the magic number of 415). Find... | 14. A. If $n \leqslant 6$, take $m=1,2, \cdots, 7$. By the pigeonhole principle, there must be a positive integer $M$ among $a_{1}, a_{2}, \cdots, a_{n}$ that is a common magic number of $i$ and $j (1 \leqslant i<j \leqslant 7)$, i.e.,
$$
7 \mid(10 M+i), 7 \mid(10 M+j) \text {. }
$$
Then $7 \mid (j-i)$. But $0<j-i \le... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x \geqslant 0$,
$f(x)=2^{x}+2 x+b$ ( $b$ is a constant).
Then $f(-10)=$ $\qquad$ . | 2. -1043 .
From the given condition, we easily know that
$$
f(0)=2^{0}+2 \times 0+b=0 \text {. }
$$
Solving for $b$ yields $b=-1$.
By the property of odd functions $f(-x)=-f(x)$, we have
$$
\begin{array}{l}
f(-10)=-f(10)=-2^{10}-2 \times 10+1 \\
=-1043 .
\end{array}
$$ | -1043 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If real numbers $x, y, z$ satisfy the equation
$$
\sqrt{x+9+\sqrt{x-7}}+\frac{|x+y-z|}{4}=4 \text {, }
$$
then the units digit of $(5 x+3 y-3 z)^{2013}$ is $\qquad$ | 3. 4 .
It is known that $x \geqslant 7$, then
$$
\left\{\begin{array}{l}
\sqrt{x+9+\sqrt{x-7}} \geqslant 4, \\
\frac{|x+y-z|}{4} \geqslant 0 .
\end{array}\right.
$$
Combining the given equations, we have
$$
\left\{\begin{array}{l}
\sqrt{x+9+\sqrt{x-7}}=4, \\
\frac{|x+y-z|}{4}=0 .
\end{array}\right.
$$
Therefore, $x=... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 3, square $ABCD$ is divided into 8 triangles of equal area. If $AG=\sqrt{50}$, then the area $S$ of square $ABCD$ is $ \qquad $. | 4. 128 .
As shown in Figure 5, draw $K L / / D C$ through point $F$, take the midpoint $N$ of $A B$, and connect $G N$ with $A H$ intersecting at point $P$.
Let the side length of the square $A B C D$ be $a$.
Given $S_{\triangle D C I}=S_{\triangle M B H}=\frac{1}{8} S$, we know
$C I=B H=\frac{1}{4} B C=\frac{a}{4}$.
... | 128 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Given real numbers $m, n$ satisfy $m-n=\sqrt{10}$, $m^{2}-3 n^{2}$ is a prime number. If the maximum value of $m^{2}-3 n^{2}$ is $a$, and the minimum value is $b$, then $a-b=$ $\qquad$ | 5.11 .
Let $m^{2}-3 n^{2}=p$ (where $p$ is a prime number).
From $m-n=\sqrt{10}$, we get
$$
m=\sqrt{10}+n \text {. }
$$
Substituting equation (2) into equation (1) and simplifying, we get
$$
\begin{array}{l}
2 n^{2}-2 \sqrt{10} n+p-10=0 \\
\Rightarrow \Delta=40-8 p+80 \geqslant 0 \\
\Rightarrow p \leqslant 15 \\
\Rig... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In the Cartesian coordinate system $x O y$, it is known that there are three points $A(a, 1), B(2, b), C(3,4)$.
If the projections of $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are the same, then $3 a-4 b=$ $\qquad$ | 3. 2 .
Solution 1 The projections of vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ in the direction of $\overrightarrow{O C}$ are $\frac{\overrightarrow{O A} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}, \frac{\overrightarrow{O B} \cdot \overrightarrow{O C}}{|\overrightarrow{O C}|}$, respectively.
A... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given three distinct integers $x, y, z$ whose sum lies between 40 and 44. If $x, y, z$ form an arithmetic sequence with a common difference of $d$, and $x+y, y+z, z+x$ form a geometric sequence with a common ratio of $q$, then $d q=$ $\qquad$ | 5. 42 .
$$
\begin{array}{l}
\text { Given } x=y-d, z=y+d \\
\Rightarrow x+y=2 y-d, y+z=2 y+d \\
\Rightarrow z+x=2 y . \\
\text { Also, }(x+y)(z+x)=(y+z)^{2} \\
\Rightarrow 2 y(2 y-d)=(2 y+d)^{2} \\
\Rightarrow d(d+6 y)=0 .
\end{array}
$$
Since $d \neq 0$, we have $d=-6 y$.
$$
\begin{array}{l}
\text { Also, } 40<x+y+z=... | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For example, $8 n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$
and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$. [4]
(2009, "Mathematics Weekly Cup" National Junior High Sch... | Let $a_{1}, a_{2}, \cdots, a_{n}$ be such that removing $a_{i} (i=1, 2, \cdots, n)$ leaves the arithmetic mean of the remaining $n-1$ numbers as a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$
Thus, for any $1 \leqslant i<j \leqslant n$, we have
$$
b_{i}-b_{j}=\f... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. If for any $x \in\left(-\frac{1}{2}, 1\right)$, we have
$$
\frac{x}{1+x-2 x^{2}}=\sum_{k=0}^{\infty} a_{k} x^{k},
$$
then $a_{3}+a_{4}=$ . $\qquad$ | 9. -2 .
In equation (1), let $x=0$, we get $a_{0}=0$. Then $\frac{1}{1+x-2 x^{2}}=\sum_{k=1}^{\infty} a_{k} x^{k-1}$. Substituting $x=0$ into the above equation, we get $a_{1}=1$. Then $\frac{-1+2 x}{1+x-2 x^{2}}=\sum_{k=2}^{\infty} a_{k} x^{k-2}$. Substituting $x=0$ into the above equation, we get $a_{2}=-1$. Similar... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. If $0 \leqslant x_{i} \leqslant 1(i=1,2, \cdots, 5)$, then
$$
M=x_{1}-x_{2}^{3}+x_{2}-x_{3}^{3}+x_{3}-x_{4}^{3}+x_{4}-x_{5}^{3}+x_{5}-x_{1}^{3}
$$
the maximum value is $\qquad$ | 10.4.
If there exists a positive integer $j$, such that
$$
x_{j}=x_{j+1}\left(j=1,2, \cdots, 5, x_{6}=x_{1}\right) \text {, }
$$
then $M \leqslant 4$.
When $x_{1}=0, x_{2}=1, x_{3}=0, x_{4}=1, x_{5}=0$, the equality holds.
If for any positive integer $i$, we have $x_{i} \neq x_{i+1}$ $(i=1,2, \cdots 5)$, then either... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. There are three sets of cards in red, yellow, and blue, each set containing five cards, marked with the letters $A, B, C, D, E$. If five cards are drawn from these 15 cards, with the requirement that the letters are all different and all three colors are included, then the number of different ways to draw the cards ... | 4. 150.
Divide into two categories: $3, 1,1$ and $2,2,1$.
Calculate respectively:
$$
\frac{C_{3}^{1} C_{5}^{3} C_{2}^{1} C_{2}^{1} C_{1}^{1}}{A_{2}^{2}}+\frac{C_{3}^{1} C_{5}^{2} C_{2}^{1} C_{3}^{2} C_{1}^{1}}{A_{2}^{2}}=150
$$ | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given a moving point $P(x, y)$ satisfies
$$
\left\{\begin{array}{l}
2 x+y \leqslant 2, \\
x \geqslant 0, \\
\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1 .
\end{array}\right.
$$
Then the area of the figure formed by the moving point $P(x, y)$ is | 8. 2 .
From equation (1), we have
$$
\begin{array}{l}
x+\sqrt{x^{2}+1} \geqslant \sqrt{y^{2}+1}-y \\
\Rightarrow \ln \left(x+\sqrt{x^{2}+1}\right) \geqslant \ln \left(\sqrt{y^{2}+1}-y\right) .
\end{array}
$$
It is easy to see that $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)$ is a strictly increasing function.
$$
\begin{a... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
14. (15 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$.
(1) Find the minimum value of $f(n)$;
(2) When $n=2 \times 10^{k}-1\left(k \in \mathbf{N}_{+}\right)$, find $f(n)$;
(3) Does there exist a positive integer $n$ such that
$$
f(n)=201... | 14. (1) Since $3 n^{2}+n+1$ is an odd number greater than 3, we know that $f(n) \neq 1$.
Assume $f(n)=2$. Then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e.,
$$
3 n^{2}+n+1=10^{k}+1 \text {. }
$$
Clearly, when $k=1$, $n$ does not exist.
Therefore, $k$ is ... | 2012 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The positive integer $n=$ $\qquad$ that makes $2^{n}+256$ a perfect square. | When $n8$, $2^{n}+256=2^{8}\left(2^{n-8}+1\right)$.
If it is a perfect square, then $2^{n-8}+1$ is the square of an odd number.
Let $2^{n-8}+1=(2 k+1)^{2}$ ( $k$ is a positive integer). Then $2^{n-10}=k(k+1)$.
Since $k$ and $k+1$ are one odd and one even, hence $k=1$.
Thus, $n=11$. | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $A=\{x \mid x \geqslant 10, x \in \mathbf{N}\}, B \subseteq A$, and the elements in $B$ satisfy:
(i) The digits of any element are all different;
(ii) The sum of any two digits of any element is not equal to 9.
(1) Find the number of two-digit and three-digit numbers in $B$;
(2) Does there exist a five-di... | (1) For two-digit numbers, the digit in the tens place can be $1,2, \cdots, 9$; the digit in the units place, since it cannot be the same as the digit in the tens place and the sum of the two digits cannot be 9, has 8 possible choices for each digit in the tens place.
Therefore, the total number of two-digit numbers th... | 4012 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 For a four-digit number, at most two of its digits are different. How many such four-digit numbers are there?
保留源文本的换行和格式，所以翻译结果如下：
Example 2 For a four-digit number, at most two of its digits are different.
Ask: How many such four-digit numbers are there? | Solution: Clearly, there are exactly 9 four-digit numbers where all four digits are the same.
Below, we consider four-digit numbers with exactly two different digits in three steps.
(1) First, consider the thousands place, which has 9 possible choices: $1,2, \cdots, 9$.
(2) Next, consider the hundreds, tens, and units... | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given $A \cup B \cup C=\{1,2, \cdots, 6\}$, and $A \cap B=\{1,2\},\{1,2,3,4\} \subseteq B \cup C$.
Then the number of $(A, B, C)$ that satisfy the conditions is $\qquad$ groups (different orders of $A, B, C$ are considered different groups). | As shown in Figure 1, for $1$ and $2$, they can belong to regions I and II, which gives $2^{2}$ possibilities; for $3$ and $4$, they can belong to $B \cup C$ except for regions I and II, i.e., regions III, IV, VI, and VII, which gives $4^{2}$ possibilities; for $5$ and $6$, they can belong to $A \cup B$ except for regi... | 1600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 (1) Find the minimum value of the sum of the digits of integers of the form $3 n^{2}+n+1\left(n \in \mathbf{Z}_{+}\right)$;
(2) Does there exist a number of this form whose sum of digits is 1999? | (1) Since
$$
3 n^{2}+n+1=n(3 n+1)+1
$$
is an odd number, the last digit must be odd.
If the sum of the digits is 1, then the number is 1, but $n$ is a positive integer, $3 n^{2}+n+1 \geqslant 5$, which is a contradiction.
If the sum of the digits is 2 and it is an odd number, then the first and last digits must both ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If 6 college graduates apply to three employers, and each employer hires at least one of them, then the number of different hiring scenarios is $\qquad$ kinds. | The number of ways to hire 3 people is $\mathrm{A}_{6}^{3}=120$; the number of ways to hire 4 people is $\frac{\mathrm{C}_{6}^{2} \mathrm{C}_{4}^{1} \mathrm{C}_{3}^{1}}{2!} \times \mathrm{A}_{3}^{3}=540$; the number of ways to hire 5 people is
$$
\frac{C_{6}^{2} C_{4}^{2} C_{2}^{1}}{2!} \times A_{3}^{3}+\frac{C_{6}^{3}... | 2100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 In a $6 \times 6$ grid, three identical red cars and three identical black cars are parked, with one car in each row and each column, and each car occupies one cell. The number of ways to park the cars is ( ).
(A) 720
(B) 20
(C) 518400
(D) 14400 | Assume first that the three red cars are distinct and the three black cars are also distinct. The first car can obviously be placed in any of the 36 squares, giving 36 ways. The second car, which cannot be in the same row or column as the first car, has 25 ways to be placed.
Similarly, the third, fourth, fifth, and si... | 14400 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
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